Rachel’s revenue function is in
If you expand and simplify it to
factored form.
factored form
express it in
then you can compare both functions to see if they
standard form,
a quadratic function in the form
f(x) 5 a(x 2 r) (x 2 s)
are the same.
standard form
a quadratic function in the form
Beth’s Solution: Comparing and Analyzing the Functions
f(x) 5 ax
1 bx 1 c
2
R(x) 5 (40 2 x) (10 1 x)
I multiplied the binomials and
collected like terms. Rachel’s revenue
R(x) 5 400 1 40x 2 10x 2 x
2
function in standard form is the same
R(x) 5 400 1 30x 2 x
2
as Andrew’s revenue function, but
the terms are in a different order.
R(x) 5 2x
1 30x 1 400
I rearranged my function to match
2
Andrew’s.
0 5 2x
1 30x 1 400
2
or
zeros of a relation
The x-intercepts, or zeros, occur
0 5 (40 2 x) (10 1 x)
R(x) 5 0.
when the revenue is 0, so
the values of x for which a
relation has the value zero. The
I used Rachel’s function to find the
40 2 x 5 0
10 1 x 5 0
and
zeros of a relation correspond to
x-intercepts by setting each factor
40 5 0 1 x
x 5 0 2 10
the x-intercepts of its graph
and
equal to zero and solving for x.
40 5 x
x 5 210
and
The maximum value is the
(40 1 (210) )
y-coordinate of the vertex, and the
x 5
2
vertex lies on the axis of symmetry.
30
To find the axis of symmetry, I added
x 5
the zeros and divided by 2.
2
x 5 15
To find the maximum revenue
R(15) 5 (40 2 15) (10 1 15)
I substituted the x-value into
Rachel’s equation.
R(15) 5 (25) (25)
R(15) 5 625
The maximum revenue is $625.
134
Chapter 3
NEL